]]>https://jpmccarthymaths.com/2019/01/11/contexts-concepts/feed/0jpmccarthymathsAnother Relative Velocity Questionhttps://jpmccarthymaths.com/2018/12/12/another-relative-velocity-question/
https://jpmccarthymaths.com/2018/12/12/another-relative-velocity-question/#respondWed, 12 Dec 2018 08:39:21 +0000http://jpmccarthymaths.com/?p=10658On a particular day the velocity of the wind, in terms of and , is , where .

and are unit vectors in the directions East and North respectively.

To a man travelling due East the wind appears to come from a direction North West where .

When he travels due North at the same speed as before, the wind appears to come from a direction North West where .

Find the actual direction of the wind.

Solution:

We start by writing

.

We have two equations.

Firstly, when the man travels due East, for some constant . We have and so

.

We know that this, with respect to the man, is coming from North West. This means we have:

And furthermore:

.

Now consider when the man travels due North, . We have still and so

.

We know that this, with respect to the man, is coming from North West. This means we have:

And furthermore:

,

but we have and so

,

so that .

This means that the velocity of the wind looks like:

That is the wind comes from North West. We have that , so the answer to the question asked is N W.

]]>https://jpmccarthymaths.com/2018/12/12/another-relative-velocity-question/feed/0jpmccarthymathsrel6.jpgrel7.jpgrel8.jpgRelative Velocity Question Three Wayshttps://jpmccarthymaths.com/2018/12/12/relative-velocity-question-three-ways/
https://jpmccarthymaths.com/2018/12/12/relative-velocity-question-three-ways/#respondWed, 12 Dec 2018 08:09:56 +0000http://jpmccarthymaths.com/?p=10651Here we present three solutions to the one problem. The vector solution is probably the slickest. The geometry solution here can be simplified by being less rigorous, and the coordinate geometry solution might be made easier by using the formula.

LCHL 2007, Q. 2(a)

Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.

(i) Find the velocity of ship A relative to ship B.

(ii) Calculate the length of time, to the nearest minute, for which the ships
are less than or equal to 8 km apart.

Solution to (i): We have that and and so

.

Vector Approach to (ii)

First of all we draw a picture. As we are talking relative to ship B we will put B at the origin. If ship B is 8 km north-east of ship A then ship A is 8 km south-west of ship B.

Where is the initial displacement of ship A relative to ship B, the displacement of ship A relative to ship B, as a function of time, is given by

.

Using some trigonometry — vertical and horizontal components of — we have

,

and so

.

We are interested in finding out for how long this displacement of ship A from ship B is less than eight:

.

This is a -quadratic and so is negative between the roots, where

.

Coordinate Geometry Approach to (ii)

We have:

Relative to B, ship A travels along a line —and the set of points a distance eight from comprise a circle, so we have:

The line through has slope equal to (why) and has as a point the initial position of A relative to B,

and so the equation of the line through is:

.

The equation of the circle is . These curves — the line and the circle — intersect at and — and we already know where is.

To find the intersection, substitute into :

.

The sensible thing here is to use the formula but this actually has nice factors. Multiply . Consider . The sum is 172 — too big — but try : bingo .

So we can rewrite

Setting the first factor equal to zero gives . The other gives

.

We can find the corresponding coordinate by substituting back into :

.

Now the distance between these points is found using the distance formula:

.

The speed of ship A relative to B is

,

and so using .

Geometric/Trigonometric Approach to (ii)

There are probably a number of geometric/trigonometric approaches. One approach is to drop a perpendicular from to — probably the easiest way. Here is another approach.

We have:

We have — and the set of points a distance eight from comprise a circle. Therefore we consider the following — ship A is headed to (relative to ship :

Now produce the line segment and the line segment . Note that and . Define .

(It is possible to get an approximate value for using — and when I saw that the answer was to be approximate (nearest minute) — I thought perhaps I should have done this instead of what follows.)

If you would like to help us more with the Transposition Project, please take this survey.

Mathematics Exam Advice

The first piece of advice is to read questions carefully. Don’t glance at a question and go off writing: take a moment to understand what you have been asked to do.

Don’t use tippex; instead draw a simple line(s) through work that you think is incorrect.

For equations, check your solution by substituting your solution into the original equation. If your answer is wrong and you know it is wrong: write that on your script.

If you do have time at the end of the exam, go through each of your answers and ask yourself:

have I answered the question that was asked?

does my answer make sense? If no, say so, and then try and fix your solution.

check your answer (e.g. if you are looking at a general true, look at a special case; substitute your solution into equations; check your answer against a rough estimate; or what a picture is telling you; etc). If your answer is wrong, say so, and then try and fix your solution.

Student Feedback

You are invited to give your feedback on my teaching and this module here.

Week 12

Week 13

We will have three review lectures, and tutorials as normal.

I have drafted a sample paper and we will go through this exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.

Tutorials on Week 13 at the usual times and venue. As long as there are not too many people in a tutorial (max 18), you can attend both tutorials if you want.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, exam papers etc.

The first piece of advice is to read questions carefully. Don’t glance at a question and go off writing: take a moment to understand what you have been asked to do.

Don’t use tippex; instead draw a simple line(s) through work that you think is incorrect.

For equations, check your solution by substituting your solution into the original equation. If your answer is wrong and you know it is wrong: write that on your script.

If you do have time at the end of the exam, go through each of your answers and ask yourself:

have I answered the question that was asked?

does my answer make sense? If no, say so, and then try and fix your solution.

check your answer (e.g. for a fitted curve or beam function, input values and see do they make sense; substitute your solution into equations; check your answer against a rough estimate; or what a picture is telling you; etc). If your answer is wrong, say so, and then try and fix your solution.

Student Feedback

You are invited to give your feedback on my teaching and this module here.

Assessment 2

I am working my way through these. I promise you your final CA results before Monday morning.

Week 12

We finished off Chapter 4 by looking at Error Analysis, including rounding error. We had a tutorial for a lecture on Wednesday, and Thursday might allow some tutorial time.

Week 13

We will go through last year’s exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.

We will also have tutorials on Friday 14 December at the usual times and venue. If there aren’t too many students present (max 18), you can attend both tutorials.

Week 12

Week 13

We will have three review lectures, and tutorials as normal.

I have drafted a sample paper and we will go through this exam on the board and then I will answer your questions if there are any. If there are none I will help one-to-one. Usual class times and locations.

Tutorials on Week 13 at the usual times and venue.

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage — especially those of us who struggled in the test.

Student Resources

Please see the Student Resources tab on the top of this page for information on the Academic Learning Centre, etc.

Week 12

On Monday and Tuesday we will have extra tutorials: you will be invited to either work on integration and/or matrices; the choice will be up to you.

On Thursday we will finish off the module by doing an extra example of a centre of gravity of a solid of revolution.

Week 13

There is an exam paper at the back of your notes — I will go through this on the board in the lecture times (in the usual venues):

Monday 16:00

Tuesday 09:00

Thursday 09:00

We will also have tutorial time in the tutorial slots. You can come to as many tutorials as you like.

Monday at 09:00 in E15

Monday at 17:00 in B189

Thursday at 12:00 in E4

Study

Please feel free to ask me questions about the exercises via email or even better on this webpage.

Student Resources

Please see the Student Resources tab on the top of this page for information on past exam papers, Academic Learning Centre, etc.

]]>https://jpmccarthymaths.com/2018/11/29/math6040-winter-2018-week-11/feed/0jpmccarthymathsExpected Number of Shared Birthdayshttps://jpmccarthymaths.com/2018/11/28/expected-number-of-shared-birthdays/
https://jpmccarthymaths.com/2018/11/28/expected-number-of-shared-birthdays/#respondWed, 28 Nov 2018 08:35:24 +0000http://jpmccarthymaths.com/?p=10607I received the following email (extract) from a colleague:

With the birthday question the chances of 23 people having unique birthdays is less than ½ so probability of shared birthdays is greater than 1-in-2.

Coincidentally on the day you sent out the paper, the following question/math fact was in my son’s 5th Class homework.

We are still debating the answer, hopefully you could clarify…

In a group of 368 people, how many should share the same birthday. There are 16×23 in 368 so there are 16 ways that 2 people should share same birthday (?) but my son pointed out, what about 3 people or 4 people etc.

I don’t think this is an easy problem at all.

First off we assume nobody is born on a leap day and the distribution of birthdays is uniform among the 365 possible birthdays. We also assume the birthdays are independent (so no twins and such).

They were probably going for 16 or 32 but that is wrong both for the reasons given by your son but also for the fact that people in different sets of 23 can also share birthdays.

The brute force way of calculating it is to call by the random variable that is the number of people who share a birthday and then the question is more or less looking for the expected value of , which is given by:

.

Already we have that (why), and is (why) the probability that four people share one birthday and 364 have different birthdays. This probability isn’t too difficult to calculate (its about ) but then things get a lot harder.

For , there are two possibilities:

5 share a birthday, 363 different birthdays, OR

2 share a birthday, 3 share a different birthday, and the remaining 363 have different birthdays

Then is already getting very complex:

6 share a birthday, 362 different birthdays, OR

3, 3, 362

4, 2, 362

2, 2, 2, 362

This problem is spiraling out of control.

There is another approach that takes advantage of the fact that expectation is linear, and the probability of an event not happening is

.

Label the 368 people by and define a random variable by

Then , the number of people who share a birthday, is given by:

,

and we can calculate, using the linearity of expectation.

.

The are not independent but the linearity of expectation holds even when the addend random variables are not independent… and each of the has the same expectation. Let be the probability that person does not share a birthday with anyone else; then

,

but , and

,

and so

.

All of the 368 have this same expectation and so

.

Now, what is the probability that nobody shares person ‘s birthday?

We need persons and — 367 persons — to have different birthdays to person , and for each there is 364/365 ways of this happening, and we do have independence here (person 1 not sharing person ‘s birthday doesn’t change the probability of person 2 not sharing person ‘s birthday), and so is the product of the probabilities.

So we have that

,

and so the answer to the question is:

.

There is possibly another way of answering this using the fact that with 368 people there are